Chapter 10.3, Problem 2GP

To determine

To Find: To find the mean absolute deviation of the given data.

Answer to Problem 2GP

The mean absolute deviation is 2.8 and 2.5

Explanation of Solution

Given information:Number of pull ups in two weeks.

Formula used: For Mean: $\overline{x}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{N}$ and for mean absolute deviation $MD=\frac{{\displaystyle \sum _{i=1}^n\left|x_i-\overline{x}\right|}}{N}$

Calculation:

Let $x_i\text{and}{y}_i$ represented the observations for the first week and second week.

Calculation of Mean

  $\begin{array}{l}\overline{x}=\frac{5+6+7+10+12+12+13}{7}\\ \overline{x}=\frac{65}{7}\\ \therefore \overline{x}=9.3\end{array}$

Similarly,

  $\begin{array}{l}\overline{y}=\frac{20+20+20+20+25+25+30}{7}\\ \overline{y}=\frac{160}{7}\\ \therefore \overline{y}=22.9\end{array}$

Thus, the average score in both the months are 104 and 125 respectively.

Now, calculation of mean absolute deviation

$x_i$	$x_i-\overline{x}$	$\left|x_i-\overline{x}\right|$	$y_i$	$y_i-\overline{y}$	$\left|y_i-\overline{y}\right|$
5	$5-9.3=-4.3$	4.3	20	$20-22.9=-2.9$	2.9
6	$6-9.3=-3.3$	3.3	20	$20-22.9=-2.9$	2.9
7	$7-9.3=-2.3$	2.3	20	$20-22.9=-2.9$	2.9
10	$10-9.3=0.7$	0.7	20	$20-22.9=-2.9$	2.9
12	$12-9.3=2.7$	2.7	25	$25-22.9=2.1$	2.1
12	$12-9.3=2.7$	2.7	25	$25-22.9=2.1$	2.1
13	$13-9.3=3.7$	3.7	30	$30-22.9=7.1$	7.1

  $\begin{array}{l}\text{MD}\text{of}x=\frac{4.3+3.3+2.3+0.7+2.7+2.7+3.7}{7}\\ \text{MD}\text{of}x=\frac{19.7}{7}\\ \text{MD}\text{of}x=2.8\end{array}$

And

  $\begin{array}{l}\text{MD}\text{of}y=\frac{2.9+2.9+2.9+2.9+2.1+2.1+7.1}{7}\\ \text{MD}\text{of}y=\frac{22.9}{9}\\ \text{MD}\text{of}y=2.5\end{array}$

Thus, mean absolute deviation is 2.8 and 2.5

Hence, the required mean absolute deviation is 2.8 and 2.5

The observations of x are more variable than the observation of y

So, y is more consistent and dependable.